The average time a person spends in each visit to an online

The average time a person spends in each visit to an online social networking service is

70 minutes.

The standard deviation is 10 minutes. If a visitor is selected at random, find the probability that he or she will spend the time shown on the networking service. Assume the times are normally distributed.

The probability that a randomly selected visitor spends at least 182 minutes per visit me is what %?

Refer to the table of values (Area Under the Standard Normal Distribution ) as needed. If necessary, round intermediate calculations to the nearest hundred

The probability that a randomly selected visitor spends at least 182 minutes per vis

Solution

The probability that a randomly selected visitor spends at least 182 minutes per visit me is what %?

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    182      
u = mean =    70      
          
s = standard deviation =    10      
          
Thus,          
          
z = (x - u) / s =    11.2      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   11.2   ) =    0 [ANSWER]

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As this is a weird answer, I think you meant that the mean is 170. In that case, if mean = 170 instead,

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    182      
u = mean =    170      
          
s = standard deviation =    10      
          
Thus,          
          
z = (x - u) / s =    1.2      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.2   ) =    0.1151 = 11.51% [ANSWER]